Question

: The nucleus of a helium atom has a charge of 2 protons. What is the magnitude of the electric field at point 1.0 × 10−10 m from the nucleus?

Answer 1

wah wah bhai wah humare level ke upar ka question

Answer 2

Given:

The nucleus of a helium atom has a charge of 2 protons.

To find:

The magnitude of the electric field at point 1.0 × 10−10 m from the nucleus?

Calculation:

Charge of each proton is + 1.6 × 10^(-19) C

So, field intensity be E:

$\therefore \: E = \dfrac{kq}{ {r}^{2} }$

$= > \: E = \dfrac{k(2 \times 1.6 \times {10}^{ - 19}) }{ {(10 \times {10}^{ - 10}) }^{2} }$

$= > \: E = \dfrac{k(2 \times 1.6 \times {10}^{ - 19}) }{ {( {10}^{ - 9}) }^{2} }$

$= > \: E = \dfrac{k(2 \times 1.6 \times {10}^{ - 19}) }{ {10}^{ - 18} }$

$= > \: E = \dfrac{k(3.2 \times {10}^{ - 19}) }{ {10}^{ - 18} }$

Putting value of Coulomb's Constant (k):

$= > \: E = \dfrac{9 \times {10}^{9} \times (3.2 \times {10}^{ - 19}) }{ {10}^{ - 18} }$

$= > \: E = \dfrac{9 \times (3.2 \times {10}^{ - 10}) }{ {10}^{ - 18} }$

$= > \: E = 28.8 \times {10}^{8} \: N/C$

So, final answer is:

$\boxed{ \bold{\: E = 28.8 \times {10}^{8} \: N/C}}$